Nitrogen dioxide decomposes according to the reaction 2NO₂(g) 2NO(g) + O₂(g)
Where K_p = 4.48 × 10^-13 at 25°C. What is the value for K_c?

Question

Nitrogen dioxide decomposes according to the reaction 2NO₂(g) 2NO(g) + O₂(g) Where K_p = 4.48 × 10^-13 at 25°C. What is the value for K_c?

Quizplus · Accepted Answer

According to the law of mass action, the equilibrium constant (Kc) for the decomposition of nitrogen dioxide can be calculated as:
Kc = [NO]^2 [O2] / [NO2]^2
Where [NO], [O2], and [NO2] are the molar concentrations of nitrogen monoxide, oxygen, and nitrogen dioxide, respectively, at equilibrium. 
But we are given the equilibrium constant (Kp) at 25°C. The relationship between Kc and Kp is:
Kp = Kc(RT)^Δn
where R is the gas constant, T is the temperature in Kelvin, and Δn is the difference in the number of moles of gas between the products and the reactants (in this case, Δn = (2 + 1) - 2 = 1).
Therefore, we can write:
Kc = Kp / (RT)
Plugging in the given values, we get:
Kc = (4.48 × 10^-13) / (0.0821 × 298)
Kc = 1.83 × 10^-14
Therefore, the answer is choice B.

Nitrogen Dioxide Decomposes According to the Reaction 2NO2(g) 2NO(g)

Nitrogen Dioxide Decomposes According to the Reaction 2NO₂(g) 2NO(g)